Algebraically closed group

In mathematics, in the realm of group theory, a group $$A is algebraically closed if any finite set of equations and inequations that "make sense" in $$A already have a solution in $$A . This idea will be made precise later in the article.

Informal discussion

Suppose we wished to find an element $$x of a group $$G satisfying the conditions (equations and inequations):

::$$x^2=1 ::$$x^3=1 ::$$x e 1

Then it is easy to see that this is impossible because the first two equations imply $$x=1 . In this case we say the set of conditions are inconsistent with $$G . (In fact this set of conditions are inconsistent with any group whatsoever.)

However if we extend the group $$G to the group $$H with multiplication table:

Then the condions have two solutions, namely $$x=b and $$x=c .

Thus there are three possibilities regarding such conditions:
* They may be inconsistent with $$G and have no solution in any extension of $$G .
* They may have a solution in $$G .
* They may have no solution in $$G but nevertheless have a solution in some extension $$H of $$G .

It is reasonable to ask whether there are any groups $$A such that whenever a set of conditions like these have a solution at all, they have a solution in $$A itself? The answer turns out to be "yes", and we call such groups algebraically closed groups.

Formal definition of an algebraically closed group

We first need some preliminary ideas.

If $$G is a group and $$F is the free group on countably many generators, then by a finite set of equations and inequations with coefficients in $$G we mean a pair of subsets $$E and $$I of $$Fstar G the free product of $$F and $$G .

This formalizes the notion of a set of equations and inequations consisting of variables $$x_i and elements $$g_j of $$G . The set $$E represents equations like: ::$$x_1^2g_1^4x_3=1::$$x_3^2g_2x_4g_1=1::$$dots The set $$I represents inequations like::$$g_5^{-1}x_3 e 1::$$dots

By a solution in $$G to this finite set of equations and inequations, we mean a homomorphism $$f:F ightarrow G, such that $$ilde{f}(e)=1 for all $$ein E and $$ilde{f}(i) e 1 for all $$iin I. Where $$ilde{f} is the unique homomorphism $$ilde{f}:Fstar G ightarrow G that equals $$f on $$F and is the identity on $$G .

This formalizes the idea of substituting elements of $$G for the variables to get true identities and inidentities. In the example the substitutions $$x_1mapsto g_6, x_3mapsto g_7 and $$x_4mapsto g_8 yield:::$$g_6^2g_1^4g_7=1::$$g_7^2g_2g_8g_1=1::$$dots ::$$g_5^{-1}g_7 e 1::$$dots

We say the finite set of equations and inequations is consistent with $$G if we can solve them in a "bigger" group $$H . More formally:

The equations and inequations are consistent with $$G if there is a group$$H and an embedding $$h:G ightarrow H such that the finite set of equations and inequations $$ilde{h}(E) and $$ilde{h}(I) has a solution in $$H . Where $$ilde{h} is the unique homomorphism $$ilde{h}:Fstar G ightarrow Fstar H that equals $$h on $$G and is the identity on $$F .

Now we formally define the group $$A to be algebraically closed if every finite set of equations and inequations that has coefficients in $$A and is consistent with $$A has a solution in $$A .

Known Results

It is difficult to give concrete examples of algebraically closed groups as the following results indicate:

* Every countable group can be embedded in a countable algebraically closed group.
* Every algebraically closed group is simple.
* No algebraically closed group is finitely generated.
* An algebraically closed group cannot be recursively presented.
* A finitely generated group has solvable word problem if and only if it can embedded in every algebraically closed group.

The proofs of these results are, in general very complex. However a sketch the proof that a countable group $$C can be embedded in an algebraically closed group follows.

First we embed $$C in a countable group $$C_1 with the property that every finite set of equations with coefficients in $$C that is consistent in $$C_1 has a solution in $$C_1 as follows:

There are only countably many finite sets of equations and inequations with coefficients in $$C . Fix an enumeration $$S_0,S_1,S_2,dots of them. Define groups $$D_0,D_1,D_2,dots inductively by:

::$$D_0 = C

::$$D_{i+1} = left{egin{matrix} D_i &mbox{if} S_i mbox{is not consistent with} D_i \langle D_i,h_1,h_2,dots,h_n angle &mbox{if} S_i mbox{has a solution in} Hsupseteq D_i mbox{with} x_jmapsto h_j 1le jle nend{matrix} ight.

Now let:

::$$C_1=cup_{i=0}^{infty}D_{i}

Now iterate this construction to get a sequence of groups $$C=C_0,C_1,C_2,dots and let:

::$$A=cup_{i=0}^{infty}C_{i}

Then $$A is a countable group containing $$C . It is algebraically closed because any finite set of equations and inequations that is consistent with $$A must have coefficients in some $$C_i and so must have a solution in $$C_{i+1} .

References

* A. Macintyre: On algebraically closed groups, ann. of Math, 96, 53-97 (1972)
* B.H. Neumann: A note on algebraically closed groups. J. London Math. Soc. 27, 227-242 (1952)
* B.H. Neumann: The isomorphism problem for algebraically closed groups. In: Word Problems, pp 553-562. Amsterdam: North-Holland 1973
* W.R. Scott: Algebraically closed groups. Proc. Amer. Math. Soc. 2, 118-121 (1951)